## anonymous 3 years ago How do I find the complex cube root of 8?

1. anonymous

method one solve $x^3=8$ $x^3-8=0$ $(x-2)(x^2+2x+4)=0$ quadratic formula will give the solutions to $x^2+2x+4=0$

2. anonymous

I see and so 2 is one root and the other 2 complex numbers will be the other 2 roots?

3. anonymous

@Satellite73

4. anonymous

method two solve $$x^3=1$$ and multiply by two since one solution is 1, divide the unit circle in two three equal parts one is at 1, the other two are $$-\frac{1}{2}+\frac{\sqrt{3}}{2}i$$ and its conjugate $$-\frac{1}{2}-\frac{\sqrt{3}}{2}i$$ multiply these numbers by 2

5. anonymous

to answer your question, yes, the solution to the quadratic will give you the two complex roots

6. anonymous

@kerstie hello

7. anonymous

ok thanks a lot, you helped me greatly

8. anonymous

yw